Question

Q1. Calculate the amount of copper produced in 1.0 hour when aqueous CuBr2 solution was electrolyzed by using a current of 4.50 A. Q2. In another electroplating experiment, if electric current was passed for 3 hours and 2.00 g of silver was deposited from a AgNO3 solution, what was the current used in amperes

Answer 1

`\boxed{\text{Q1. 3.6 g; Q2. 0.2 A}}`

Step-by-step explanation:

Q1. Mass of Cu

(a) Write the equation for the half-reaction.

Cu²⁺ + 2e⁻ ⟶ Cu

The number of electrons transferred (z) is 2 mol per mole of Cu.

(b) Calculate the number of coulombs

q = It

`\text{t} = \text{1.0 h} * \frac{\text{3600 s}}{\text{1 h}} = \text{3600 s}\\\\q = \text{3 C/s} * \text{ 3600 s} = \textbf{10 800 C}`

(c) Mass of Cu

We can summarize Faraday's laws of electrolysis as

`\begin{array}{rcl}m &=& (qM)/(zF)\\\\& = &(10 800 * 63.55)/(2 * 96 485)\\\\& = & \textbf{3.6 g}\\\end{array}\\\text{The mass of Cu produced is $\boxed{\textbf{3.6 g}}$}`

Note: The answer can have only two significant figures because that is all you gave for the time.

Q2. Current used

(a) Write the equation for the half-reaction.

Ag⁺ + e⁻ ⟶ Ag

The number of electrons transferred (z) is 1 mol per mole of Ag.

(a) Calculate q

`\begin{array}{rcl}m &=& (qM)/(zF)\\\\2.00& = &(q * 107.87)/(1 * 96 485)\\\\q &=& (2.00 * 96485)/(107.87)\\\\& = & \textbf{1789 C}\\\end{array}`

(b) Calculate the current

t = 3 h = 3 × 3600 s = 10 800 s

`\begin{array}{rcl}q&=& It\\1789 & = & I * 10800\\I & = & (1789)/(10800)\\\\& = & \textbf{0.2 A}\\\end{array}\\\text{The current used was $\large \boxed{\textbf{0.2 A}}$}`

Note: The answer can have only one significant figure because that is all you gave for the time.